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# Consider the following reaction: $2NH_3(g)+CO_2(g)\rightarrow (NH_2)_2CO(aq)+H_2O(l)$. If $637.2\;g$ $NH_3$ and $1142\;g$ $CO_2$ are allowed to react, then which one is the limiting reagent ?

$(a)\;NH_3\qquad(b)\;CO_2\qquad(c)\;either\;may\;be\qquad(d)\;none\;of\;these$

Answer: $NH_3$
Molar mass of $NH_3$ is 17.03g and that of $CO_2$ is 44.01g.
$\Rightarrow$ moles of $NH_3 = 637.2 g\;NH_3 \times \large\frac{1 mol \;NH_3}{17.03g\; NH_3}$$= 37.4 \; mol \; NH_3 \Rightarrow moles of CO_2 = 1142 g\;CO_2 \times \large\frac{1 mol \;NH_3}{44.01 g CO_2}$$ = 25.95\;mol\; CO_2$
We then calculate the number of moles of $(NH_2)_2 CO$ produced from $NH_3$ and $CO_2$
From $37.4 \; mol \; NH_3$ we get $37.4 \; mol \; NH_3 \times \large\frac{ 1 mol \; (NH_2)_2CO_2}{2 mol \; NH_3}$$= 18.71 mol \; (NH_2)_2CO_2 From 25.95 \; mol \; CO_2 we get 25.95 \; mol \; CO_2 \times \large\frac{ 1 mol \; (NH_2)_2CO_2}{1 mol \; CO_2}$$ = 25.95 mol \; (NH_2)_2CO_2$
$\Rightarrow NH_3$ must be the limiting agent as it produces a smaller amount of $(NH_2)_2CO_2$